Consider $90$ iid geometric random variables having parameter $\theta$
(and thus mean $1/\theta$), and a UMP test of the null hypothesis that
$\theta \leq 0.1$ against the alternative that $\theta \gt 0.1$. Give the
approximate p-value (based on large sample results) which results for
the case of the sum of the observations equal to $723$.

This is a practice problem (not homework). I'm given that the answer should be $0.025$.

I have that if $X_i$ are iid random variables from a full one-parameter exponential family distribution, then for a UMP test of

$$H_0:\theta\leq\theta_0 \text{ vs. } H_1:\theta\gt\theta_0$$

we can use

$$T\left(\vec{X}\right)=\sum_{i=1}^n t(X_i)$$

as a test statistic.

We have that the geometric distribution is a full one-parameter exponential family with

Consider the hypotheses $H_0: \theta \leqslant \theta_0$ and $H_A: \theta > \theta_0$. For these hypotheses, smaller values of the sample mean are more conducive to the alternative hypothesis, and so you have p-value:

In your calculation you made an error in the last line where you substituted $T(\vec{X}) = 723$ instead of $T(\vec{X}) = 723-90 = 633$ (consistent with your definition of $T(\vec{X})$). If you correct this error then you get the same value as in my working.